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A227499
Number of the Lipschitz quaternions in a reduced system modulo n.
16
1, 8, 48, 128, 480, 384, 2016, 2048, 3888, 3840, 13200, 6144, 26208, 16128, 23040, 32768, 78336, 31104, 123120, 61440, 96768, 105600, 267168, 98304, 300000, 209664, 314928, 258048, 682080, 184320, 892800, 524288, 633600, 626688, 967680, 497664, 1822176
OFFSET
1,2
LINKS
Catalina Calderón, Jose Maria Grau, A. Oller-Marcén, and László Tóth, Counting invertible sums of squares modulo n and a new generalization of Euler's totient function, Publicationes Mathematicae-Debrecen, Vol. 87 (1-2) (2015), pp. 133-145; arXiv preprint, arXiv:1403.7878 [math.NT], 2014.
FORMULA
Multiplicative: a(2^s) = 2^(4s-1); a(3^s) = 16*3^(4s-3); a(5^s) = 32*3*5^(4s-3).
From Amiram Eldar, Feb 13 2024: (Start)
Multiplicative with a(2^e) = 2^(4*e-1), and a(p^e) = p^(4*e-3) * (p-1)^2 * (p+1) for an odd prime p.
Dirichlet g.f.: zeta(s-4) * (1 - 1/2^(s-3)) * Product_{p prime > 2} (1 - 1/p^(s-3) - (p-1)/p^(s-1)).
Sum_{k=1..n} a(k) = (12/55) * c * n^5 + O(n^4 * log(n)), where c = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.53589615382833799980... (Calderón et al., 2015).
Sum_{n>=1} 1/a(n) = (17*Pi^8/57240) * Product_{p prime} (1 - 2/p^2 + 1/p^4 + 1/p^5 + 2/p^6 - 1/p^8) = 1.16039588611967540703... . (End)
MATHEMATICA
cuater[n_] := Flatten[Table[{a, b, c, d}, {a, n}, {b, n}, {c, n}, {d, n}], 3]; a[n_] := Length@Select[cuater[n], GCD[#.#, n] == 1 &]; Array[a, 20]
f[p_, e_] := (p-1)*p^(4*e-1) * If[p == 2, 1, 1 - 1/p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 13 2024 *)
PROG
(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; (p-1)*p^(4*e-1) * if(p == 2, 1, 1 - 1/p^2)); } \\ Amiram Eldar, Feb 13 2024
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
STATUS
approved